# Derivation of Darcy's law from Stokes equation

On the Wikipedia entry of Darcy's law, a derivation of Darcy's law from Stokes equation is provided. The derivation starts at the Stokes equation, which reads:

$$\mu \nabla^2 u_i + \rho g_i - \partial_i p = 0$$

where $$\mu$$ is the viscosity, $$u$$ the flow velocity, $$\rho$$ the fluid density, $$g$$ acceleration due to gravity, $$p$$ the fluid pressure, and $$\partial$$ denotes the partial derivative, all taken in the $$i$$-th direction ($$x$$, $$y$$, $$z$$, etc.). It is then said that:

Assuming the viscous resisting force is linear with the velocity we may write: $$- \left(k_{ij} \right)^{-1} \mu \phi u_j + \rho g_i - \partial_i p = 0$$

I fail to see how this assumption of leads to $$\nabla^2 u_i = - \left(k_{ij} \right)^{-1} \phi u_j.$$ Would someone be so kind to explain this step in more detail?

I don't like the derivation in the Wiki article because it leaves out all the steps. To get their equation, what you can do is assume a continuous distribution of pore number density and pore diameter with pore direction. When you integrate over these pore direction distributions, you end up with an anisotropic permeability tensor.

• Assuming that $k_{ij} = k I_{ij}$ (i.e. $k$ is isotropic), could you show me how $\nabla^2 u_i$ essentially reduces to $u_i$ in this case?
– MPA
Nov 4, 2018 at 8:39
• For a pore pointing in any arbitrary direction, the Hagen-Poiseulle equation applies within the pore (assuming constant pore diameter). The equation gives the pressure gradient in a tube in proportion to the mean velocity of flow in the tube. This is the solution to your differential equation for the tube. Nov 4, 2018 at 12:22